Another Maths Problem

[nitrocan]

I am just finished with this problem, and I wanted to see if there are other possible answers to this problem:

Show that this equation has no solutions when x and y are integers (im not so sure about the English terminology for maths, x and y must be positive or negative integers)

I will post my solution after some time.

 

[Tim Reynolds]

Assume the equation has a solution (x,y). Consider the equation modulo 5 (that is, consider the remainder of each term when divided by 5). Since x^2 is an integer, 15x^2 is divisible by 5, so it's 0 (mod 5). y^2 can be either 0, 1, or 4 (mod 5), so 7y^2 can be either 0, 2, or 3 (mod 5). 9 is equivalent to 4 mod 5. So you have 0 - (one of 0, 2, 3) = 4 (mod 5), but none of 0, 2, or 3 satisfy that equation, so we have a contradiction. Thus no solution can exist.

 

[nitrocan]

oh my god this is the exact solution that i had found. but in the mean time i found another one:
15 and 9 are divisible by 3, so 7y^2 should be divisible to 3 for the equation to have a solution. Then we can say that y = 3t
The equation becomes: 15x^2 - 7.9t^2 = 9 and when we divide the equation to 3, it becomes: 5x^2 - 21t^2 = 3 and from that, we can see that x has to be divisible by 3 as well. x = 3s, then the equation becomes
5.9s^2 - 21t^2 = 3 => 15s^2 - 7t^2 = 1
if we consider the equation in modulo 3 now, it becomes 2t^2=4
t^2 = 2(mod 3), but t^2 = 3n + 2 has no integer solutions.